What is the integral of (x^4-5x^2+3x-4)/(x+1) ?

The integral of (x^4-5x^2+3x-4)/(x+1) is (x^4)/4-(x^3)/3+(x^2)/2+12x-11ln|x+1|-5/2 (x+1)^2+13+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^4-5x^2+3x-4)/(x+1)

\int \frac{x ^{4} - 5 x ^{2} + 3 x - 4}{x + 1} d x

\frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + 12 x - 11 ln ∣ x + 1 ∣ - \frac{5}{2} (x + 1)^{2} + 13 + C

Solution steps

\int \frac{x ^{4} - 5 x ^{2} + 3 x - 4}{x + 1} d x

Expand

\frac{x ^{4} - 5 x ^{2} + 3 x - 4}{x + 1} : \frac{x ^{4}}{x + 1} - \frac{5 x ^{2}}{x + 1} + \frac{3 x}{x + 1} - \frac{4}{x + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{4}}{x + 1} d x - \int \frac{5 x ^{2}}{x + 1} d x + \int \frac{3 x}{x + 1} d x - \int \frac{4}{x + 1} d x

\int \frac{x ^{4}}{x + 1} d x = \frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - x + ln ∣ x + 1 ∣

\int \frac{5 x ^{2}}{x + 1} d x = 5 (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣)

\int \frac{3 x}{x + 1} d x = 3 (x + 1 - ln ∣ x + 1 ∣)

\int \frac{4}{x + 1} d x = 4 ln ∣ x + 1 ∣

= \frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - x + ln ∣ x + 1 ∣ - 5 (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣) + 3 (x + 1 - ln ∣ x + 1 ∣) - 4 ln ∣ x + 1 ∣

Simplify

\frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - x + ln ∣ x + 1 ∣ - 5 (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣) + 3 (x + 1 - ln ∣ x + 1 ∣) - 4 ln ∣ x + 1 ∣ : \frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + 12 x - 11 ln ∣ x + 1 ∣ - \frac{5}{2} (x + 1)^{2} + 13

= \frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + 12 x - 11 ln ∣ x + 1 ∣ - \frac{5}{2} (x + 1)^{2} + 13

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + 12 x - 11 ln ∣ x + 1 ∣ - \frac{5}{2} (x + 1)^{2} + 13 + C

\int 3 - y d y

\int (x - 3 x) d x

\int 26 x d x

\int \frac{2 x}{( x + 2 ) ( x + 3 )} d x

\int \frac{x ^{4}}{( x - 1 ) ^{2}} d x

What is the integral of (x^4-5x^2+3x-4)/(x+1) ?
The integral of (x^4-5x^2+3x-4)/(x+1) is (x^4)/4-(x^3)/3+(x^2)/2+12x-11ln|x+1|-5/2 (x+1)^2+13+C